In a triangle, an altitude is a segment of the line through a vertex perpendicular to the opposite side. An altitude is the portion of the line between the vertex and the foot of the perpendicular. Using the standard notations, in ΔABC, there are three altitudes: AH_a, BH_b, CH_c. The three lines meet at a point - the orthocenter of the triangle. For an obtuse triangle (having one angle exceeding 90^o), the orthocenter lies outside the triangle, and the segments AH_a, BH_b, CH_c do not meet. However, their extended lines do, so even in this case, it is common to assert that the altitudes are concurrent, that is pass through a point.

Let's observe that, if H is the orthocenter of ΔABC, then A is the orthocenter of ΔBCH, while B and C are the orthocenters of triangles ACH and ABH, respectively.

I have collected several proofs of the concurrency of the altitudes, but of course the altitudes have plenty of other properties not mentioned below. For example, due to the mirror property the orthic triangle solves Fagnano's Problem. The foot of an altitude also has interesting properties.

1. Altitudes as Cevians
   

   This is Corollary 3 of Ceva's theorem.

2. Orthocenter as Circumcenter
   

   The orthocenter of ΔABC coincides with the circumcenter of ΔA'B'C' whose sides are parallel to those of ΔABC and pass through the vertices of the latter.

3. Orthocenter as the Isogonal Conjugate of the Circumcenter

   Besides being the bisector of angle A, l_a also bisects the angle formed by h_a and the diameter of the circumscribed circle that contains A. It follows that that diameter and h_a are isogonal images of each other. The same is true for the vertices B and C. Therefore, H is the isogonal conjugate of the circumcenter O.

4. Orthocenter as Incenter
   

   In ΔABC, ΔH_aH_bH_c is known as the orthic triangle. It has an interesting property that its angle bisectors serve in fact as altitudes of ΔABC. Thus, the fact that, in a triangle, angle bisectors are concurrent, implies the fact that altitudes in a triangle are also concurrent.

   

   In the proof I shall repeatedly use Euclid's Proposition III.21 about inscribed angles and its reverse. Angles BH_cC, AH_aB, AH_aC, BH_bC are all right. Therefore, we get three quadrilaterals inscribable in a circle: BH_cHH_a, BH_cH_bC, and CH_bHH_a. In each, there is a pair of equal angles. Respectively: ∠H_cBH =  ∠H_cH_aH, ∠H_cBH_b =  ∠H_cCH_b, and ∠H_bCH =  ∠H_bH_aH. It remains only to note that, naturally, ∠H_cBH =  ∠H_cBH_b and ∠H_bCH =  ∠H_bCH_c. Finally, ∠H_cH_aH = ∠H_bH_aH, which proves that H_aH is an angle bisector in the orthic triangle. The other two angles are treated similarly.

5. Via the Euler Line
   

   The argument that shows that three points - the circumcenter O, the centroid M, and the orthocenter H - lie on the same line is reversible.

   Indeed, in ΔABC consider the centroid M and the circumcenter O. If they coincide, then so are the corresponding medians and the perpendicular bisectors. In other words, the medians are perpendicular to the sides and, therefore, coincide with the altitudes. The altitudes then intersect at the centroid of the triangle (which is obviously equilateral in this case.)

   Assume that the points O and M are distinct. They define a unique line on which we'll consider a point, denoted as H, such that MH = 2·OM with M lying between O and H. Since also AM = 2·MM_a, ΔAHM is similar to ΔM_aOM. Elements VI.2 implies that lines OM_a and AH are parallel. But the former is perpendicular to BC and, therefore, so is the latter. Similarly, BH  AC and CH  AB.

6. Complex Variables
   

   A proof in the circular coordinates leads directly to the Euler line and a nice theorem by J.L.Coolidge

7. Complex Variables II
   

   Two short proofs of which the second is the clearest proof I ever came across.

8. Vector Algebra I
   

   Given ΔABC, select any point O as the origin and consider vectors OA, OB, and OC that start at O and end at the vertices of the triangle. Introduce "side" vectors: AB = OB - OA, BC = OC - OB, and AC = OC - OA. In a similar manner, other vectors will be used that lie along straight lines associated with the triangle. Assume H is the point of intersection of AH_a and BH_b. Then AH  BC and BH  AC. The scalar product of orthogonal vectors is 0. We thus have two equations

   (OH - OA).(OC - OB) = 0 and (OH - OB).(OC - OA) = 0

   Subtract the first equation from the second, multiply out and simplify:

   OH.OB + OA.OC - OB.OC - OH.OA = (OH - OC).(OB - OA) = CH.AB = 0

   Therefore CH  AB. Thus the third altitude CH_c passes through the point of intersection of the first two.

9. Vector Algebra II
   

   Let now O be the circumcenter of ΔABC. Define H via OH = OA + OB + OC. We are going to show that H lies on each of the altitudes of ΔABC. For example,

   AH.BC = (AO + OH).(BO + OC)   = (-OA + OA + OB + OC).(OC - OB)   = (OB + OC).(OC - OB)   = OC.OC - OB.OB   = OC2 - OB2   = 0,

   because O is the circumcenter of ΔABC.

10. Elementary Geometry, Inscribed Angles
    

    Thanks to Bianco for this proof. See also Altshiller-Court's College Geometry, p. 94.

    Let H be the point of intersection of two altitudes BH_b and CH_c. We are going to prove that the line AH is perpendicular to BC.

    The quadrilateral CH_bH_cB is cyclic. Indeed, since the angles at H_b and H_c are right, the quadrilateral is inscribed in the circle with diameter on BC. From here, ∠BCH_c = ∠BH_bH_c. On the hand, the quadrilateral AH_bHH_c is also quadrilateral, as the circle with diameter AH passes through all four points. Therefore, ∠HH_bH_c = ∠HAH_c. Combining the two equalities, we get ∠BCH_c = ∠HAH_c.

    Extend AH beyond H. Let G be the point of intersection of AH and BC. In triangles CHG and AHH_c, ∠GCH = ∠HAH_c and also ∠CHG = ∠AHH_c. The triangles are therefore similar. Which implies that ∠HGC = ∠HH_cA = 90^o making CG the third altitude.

11. Plain Analytic Geometry
    

    (Vladimir Zajic.) Assume a triangle ABC in a carthesian coordinate system. Assume that no side is parallel to any of the 2 coordinate axes (x, y). If yes, we can always rotate the coordinate system by an arbitrary angle different from all triangle interior angles. Since the coordinate axes x, y are perpendicular to each other and since each altitude is perpendicular to one side, it follows that no altitude is parallel to any coordinate axis either. Let the coordinates of the 3 vertices be:

    A = (xA, yA) B = (xB, yB) C = (xC, yC).

    Equations of the 3 side lines are

    c = AB: y - yA = {(yA - yB)· x + xA· yB - xB· yA} / (xA - xB) a = BC: y - yB = {(yB - yC)· x + xB· yC - xC· yB} / (xB - xC) b = CA: y - yC = {(yC - yA)· x + xC· yA - xA· yC} / (xC - xA)

    We have to calculate only 1 equation, the other 2 are given by cyclic permutation of indices A, B, C.

    Lemma

    Two lines (none parralel to any coordinate axis) are perpendicular to each other if and only if the product of their tangents is equal to -1 (minus one).

    Equations of the 3 altitudes CH_c, BH_c, AH_a are obtained by using the tangents of the side lines, the lemma, and the fact that they pass through the corresponding vertex. Again, we have to set up only 1 equation, the other 2 are given by the cyclic permutation of A, B, C.

    CHc: y - yC = {-(xA - xB)· x + (xA - xB)· xC} / (yA - yB)

    AHa: y - yA = {-(xB - xC)· x + (xB - xC)· xA} / (yB - yC) BHb: y - yB = {-(xC - xA)· x + (xC - xA)· xB} / (yC - yA)

    To find the coordinates of the intersect (the orthocenter), take any two altitude equations and solve for x and y. For example, CH_c x BH_b:

    xO = {xA· xB· (yA - yB) + xB· xC· (yB - yC) + xC· xA· (yC - yA) - (yA - yB)· (yB - yC)· (yC - yA)} / (xC· yB - xB· yC + xA· yC - xC· yA + xB· yA - xA· yB)

    yO = {yA· yB· (xA - xB) + yB· yC· (xB - xC) + yC· yA· (xC - xA) - (xA - xB)· (xB - xC)· (xC - xA)} / (yC· xB - yB· xC + yA· xC - yC· xA + yB· xA - yA· xB)

    Since the solution is invariant with respect to the cyclic permutation of A, B, C, it follows that the same coordinates x_O and y_O are solution of any two altitude coordinates and the 3 altitudes indeed intersect in a single point. This could be also verified by a direct solution of all altitude equation pairs.

12. Plane Geometry
    

    (Vladimir Zajic.) Assume a ΔABC, the side c = AB is horizontal, the vertex C is above. Extend the sides a = BC and b = CA upwards beyond the vertex C.

    Construct the altitudes h_a and h_b by dropping normals from the vertices A and B to the opposite sides a = BC and b = CA, respectively. Denote O the intersection of these two altitudes. Denote H_a and H_b the heels of the altitudes h_a and h_b, respectively (i.e., their intersections with the corresponding triangle sides).

    Construct the altitude h_c of the ΔABO by dropping a normal from the vertex O to the side c = AB. Denote H_c the heel of this altitude. Extend the altitude h_c of the ΔABO upwards until it intersects both the (extended) lines a = BC and b = CA. Suppose that these intersections might be different from each other (see the attached drawing). Denote the intersections C_a and C_b, respectively. Then either OC_a < OC_b or OC_a > OC_b.

    Note that the following triangle pairs are similar (because both triangles in each pair have the same angles at the vertex O and each triangle has a right angle):

    ΔAOHb and ΔBOHa

    ΔAOHc and ΔCaOHa

    ΔBOHc and ΔCbOHb

    Consequently

    OA/OHb = OB/OHa

    OA/OHc = OCa/OHa

    OB/OHc = OCb/OHb

    Eliminating OA and OH_a by dividing the left sides and the right sides of the first two equations we get

    OHc/OHb = OB/OCa

    Multiplying the left sides and the right sides of the result and of the third equation eliminates almost everything else:

    OB/OHb = OCb/OHb· OB/OCa

    OCb/OCa = 1

    The line segments OC_b and OC_a are equal. In other words, the lines a and b intersect on the normal h_c to the line c dropped from the point O and the points C, C_b, and C_a are identical. Q.E.D.

13. Altitudes As Radical Axes
    

    (Vladimir Zajic.) Form three circles C_a, C_b, and C_c on the sides BC, AC, and AB of ΔABC as diameters. Circles C_a and C_b meet at C and one other point. This point lies on AB and is in fact the foot H_c of the altitude CH_c. Indeed, let K be the point of intersection (other than B) of C_a with AB. Then CKB = 90^o, so that CK is perpendicular to AB. Therefore K = H_c. Similarly, circle C_b meets AB at H_c (and, of course, A.) We conclude that two circles C_a and C_b that obviously meet at C, also meet at H_c. CH_c is therefore the radical axis of the two circles.

    Turning to other sides and pairs of circles, we see that the altitudes of ΔABC serve as radical axes of the circles C_a, C_b, and C_c taken in pairs. As we know, the pairwise radical axes of three circles concur in a point, and so do the three altitudes of a triangle.

    (A more general statement appears as Theorem 184 in A Treatise On the Circle and the Sphere by J. L. Coolidge: The orthocenter of a triangle is the radical center of any three circles each of which has a diameter whose extremities are a vertex and a point on the opposite side line, but no two passing through the same vertex. The proof is practically the same.)

14. Via Carnot's Theorem
    

    The identity

    (1)	AC'2 - BC'2 + BA'2 - CA'2 + CB'2 - AB'2 = 0.

    in Carnot's theorem is easily verified when AA', BB', CC' are the altitudes of ΔABC.

References

1. N. Altshiller-Court, College Geometry, Dover, 1980
2. J. L. Coolidge, A Treatise On the Circle and the Sphere, AMS - Chelsea Publishing, 1971
3. H.S.M. Coxeter, S.L. Greitzer, Geometry Revisited, MAA, 1967
4. H. S. M. Coxeter, Introduction to Geometry, John Wiley & Sons, 1961
5. D. Pedoe, Geometry: A Comprehensive Course, Dover, 1970
6. V. V. Prasolov, Essays on Numbers and Figures, AMS, 2000

Copyright © 1996-2008 Alexander Bogomolny